use Kirchhoff’s laws to solve simple circuit problems

Objective

Use Kirchhoff’s laws to analyse and solve simple electrical circuit problems.

1. Introduction

Kirchhoff’s laws are fundamental tools for analysing circuits that contain more than one loop or junction. They are based on the conservation of charge and energy.

1.1 Kirchhoff’s Current Law (KCL)

KCL states that the algebraic sum of currents meeting at a junction is zero:

$$\sum_{k=1}^{n} I_k = 0$$

In practice this means that the total current entering a node equals the total current leaving the node.

1.2 Kirchhoff’s \cdot oltage Law (K \cdot L)

K \cdot L states that the algebraic sum of the potential differences (voltages) around any closed loop is zero:

$$\sum_{k=1}^{m} V_k = 0$$

When moving around a loop, a rise in potential (e.g., across a battery from – to +) is taken as positive, and a drop (e.g., across a resistor in the direction of current) is taken as negative.

2. Applying the Laws

1. Identify all independent loops and junctions in the circuit.
2. Assign a current direction to each branch (choose arbitrarily; a negative result indicates the opposite direction).
3. Write KCL equations for each junction (except one, which will be dependent).
4. Write K \cdot L equations for each independent loop.
5. Solve the simultaneous equations for the unknown currents or voltages.

3. Symbol Table

Symbol	Quantity	Unit
$I$	Current	A (ampere)
$V$	Potential difference (voltage)	V (volt)
$R$	Resistance	Ω (ohm)
$\mathcal{E}$	Electromotive force (battery emf)	V (volt)

4. Example Problem

Problem: In the circuit below, a 12 V battery is connected to three resistors: $R_1 = 2\ \Omega$, $R_2 = 3\ \Omega$, and $R_3 = 4\ \Omega$. $R_1$ and $R_2$ are in parallel, and this combination is in series with $R_3$. Find the current through each resistor.

Solution

1. Assign currents:
   • $I$ – total current from the battery.
   • $I_1$ – current through $R_1$.
   • $I_2$ – current through $R_2$.
2. Apply KCL at the junction where the parallel branches split: $$I = I_1 + I_2$$
3. Apply K \cdot L to the left loop (battery → $R_3$ → $R_1$ → back): $$12 - I\,R_3 - I_1\,R_1 = 0$$
4. Apply K \cdot L to the right loop (battery → $R_3$ → $R_2$ → back): $$12 - I\,R_3 - I_2\,R_2 = 0$$
5. Substitute the resistor values:
   • KCL: $I = I_1 + I_2$
   • K \cdot L₁: $12 - 4I - 2I_1 = 0$ → $2I_1 = 12 - 4I$ → $I_1 = 6 - 2I$
   • K \cdot L₂: $12 - 4I - 3I_2 = 0$ → $3I_2 = 12 - 4I$ → $I_2 = 4 - \tfrac{4}{3}I$
6. Insert $I_1$ and $I_2$ into the KCL equation: $$I = (6 - 2I) + \left(4 - \frac{4}{3}I\right)$$ $$I = 10 - \frac{10}{3}I$$ $$I + \frac{10}{3}I = 10$$ $$\frac{13}{3}I = 10$$ $$I = \frac{30}{13}\ \text{A} \approx 2.31\ \text{A}$$
7. Find $I_1$ and $I_2$: $$I_1 = 6 - 2I = 6 - 2\left(\frac{30}{13}\right) = \frac{48}{13}\ \text{A} \approx 3.69\ \text{A}$$ $$I_2 = 4 - \frac{4}{3}I = 4 - \frac{4}{3}\left(\frac{30}{13}\right) = \frac{22}{13}\ \text{A} \approx 1.69\ \text{A}$$

5. Practice Questions

1. In a circuit, a 9 V battery supplies two resistors $R_1 = 5\ \Omega$ and $R_2 = 10\ \Omega$ connected in series with a third resistor $R_3 = 15\ \Omega$ placed in parallel with $R_2$. Determine the current through each resistor.
2. A circuit contains a 6 V battery and three resistors: $R_A = 2\ \Omega$, $R_B = 4\ \Omega$, and $R_C = 6\ \Omega$. $R_A$ and $R_B$ are in series, and this series combination is in parallel with $R_C$. Use Kirchhoff’s laws to find the voltage across each resistor.
3. For the network shown (battery $E = 12\ \text{V}$, resistors $R_1 = 3\ \Omega$, $R_2 = 6\ \Omega$, $R_3 = 2\ \Omega$ forming a triangle), assign currents $I_1$, $I_2$, $I_3$ clockwise in each branch and write the three K \cdot L equations.

6. Summary

• KCL conserves charge at junctions: total current in = total current out.
• K \cdot L conserves energy around loops: sum of voltage rises = sum of voltage drops.
• Systematic application of the two laws yields a set of linear equations that can be solved for unknown currents or voltages.
• Choosing current directions arbitrarily is acceptable; a negative solution simply reverses the assumed direction.